(1)W=-P V=-2.23KJ U=0 Q=-W=2.23KJ S=(2) Q=0, U W=-1.59KJ, S 10.44J mol 1k 1
2.解:设计过程如下:
Sm Hm
5℃ 100kpa H2O(l) 5℃ 100kpa H2O(s)
Q
=7.48J mol 1k 1
T
S 1
H1
3
S2 H2
0℃ 100kpa H2O(l) 0℃ 100kpa H2O(s)
273.15268.15
Hm H1 H2 H3 5820.5J mol 1
268.15
Cp,m(H2O,l)dT+(- fusHm(H2O))+
273.15
Cp,m(H2O,s)dT
Sm S1 S2 S3=nCp,m(H2O,l)ln Gm Hm T Sm 108.91J mol 1
3. 解
T2 fusHm(H2O)T++nCp,m(H2O,s)ln1 21.3J k 1T1268.15T2
G
Hdlnk H(1) G= H-T S,当T足够大时, G<0,()P 2 =2>0 T ,K ,a
TTdTT
a2
(2) k P/P P ,a ,P=9.37kpa2
1 a
4.解
(1):k ln (2):
1
t1
0.0193min 1 1 a
dcdc1
|t 0 5.79 10 3moll1 min|t 0 3.94 10 3moll 1min 1 , dtdt
(3):水解率a=0.538 ,ta 0.60=47.48min
(4):207.7L
四 选做题 1. 解:
图形如下:要提纯,设组成为a的液相,降温至b,分离液相与纯B,液相继续降温再分离纯B,如图中虚线所示。各相区的自由度数分别为:L:F=2 A+L:F=1 B+L:F=1 A+B:F=1
a